Question

The differential equation of $y = Ae^{2x} + Be^{-2x}$ is :

• A. $\frac{dy}{dx} - 4y = 0$
• B. $\frac{d^2 y}{dx^2 } - 4y = 0$
• C. $\frac{d^2y}{dx^2} = y^2$
• D. $\frac{d^2y}{dx^2} - y = 0$

Answer 1

Answer: (B)

$\frac{d^2 y}{dx^2 } - 4y = 0$

Answer 2

Given : $y = Ae^{2x} + Be^{-2x}$ $\therefore \frac{dy}{dx} = Ae^{2x} . 2 + Be^{-2x} \left(-2\right)$ $\Rightarrow \frac{dy}{dx} = 2 Ae^{2x} - 2 Be^{-2x}$ $\frac{d^{2}y}{dx^{2} } = 2Ae^{2x}.2- 2 Be^{-2x} (- 2)$ $= 4 Ae^{2x} + 4 Be^{-2x}$ $\Rightarrow \frac{d^{2}y}{dx^{2} } = 4 y \Rightarrow \frac{d^{2}y}{dx^{2}} - 4y = 0$